NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

These Solutions are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 8
Chapter Name	Application of Integrals
Exercise	Ex 8.1, Ex 8.2
Number of Questions Solved	20
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

Chapter 8 Application of Integrals Exercise 8.1

Question 1.
Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4, and the x-axis.
Solution:
The curve y² = x is a parabola with vertex at origin.Axis of x is the line of symmetry, which is the axis of parabola. The area of the region bounded by the curve, x = 1, x=4 and the x-axis. Area LMQP

Question 2.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and x-axis in the first quadrant
Solution:
The given curve is y² = 9x, which is a parabola with vertex at (0, 0) and axis along x-axis. It is symmetrical about x-axis, as it contains only even powers of y. x = 2 and x = 4 are straight lines parallel toy-axis at a positive distance of 2 and 4 units from it respectively.
∴ Required area = Area ABCD

Question 3.
Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Solution:
The given curve x² = 4y is a parabola with vertex at (0,0). Also since it contains only even powers of x,it is symmetrical about y-axis.y = 2 and y = 4 are straight lines parallel to x-axis at a positive distance of 2 and 4 from it respectively.

Question 4.
Find the area of the region bounded by the ellipse
Solution:
The equation of the ellipse is
The given ellipse is symmetrical about both axis as it contains only even powers of y and x.

Question 5.
Find the area of the region bounded by the ellipse
Solution:
It is an ellipse with centre (0,0) and length of major axis = 2a = 2×3 = 6 and length of minor axis = 2b = 2 × 2 = 4.

Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x² + y² = 4. .
Solution:
Consider the two equations x² + y² = 4 … (i)
and x = √3y i.e. …(ii)
(1) x² + y² = 4 is a circle with centre O (0,0) and radius = 2.

Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line
Solution:
The equation of the given curve are
x² + y² = a² …(i) and …(ii)
Clearly, (i) represent a circle and (ii) is the equation of a straight line parallel to y-axis at a

Question 8.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Solution:
Graph of the curve x = y² is a parabola as given in the figure. Its vertex is O and axis is x-axis. QR is the ordinate along x = 4

Question 9.
Find the area of the region bounded by the parabola y = x² and y = |x|.
Solution:
Clearly x² = y represents a parabola with vertex at (0, 0) positive direction of y-axis as its axis it opens upwards.
y = |x| i.e., y = x and y = -x represent two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.
The required region is the shaded region as shown in the figure. Since both the curve are symmetrical about y-axis. So,
Required area = 2 (shaded area in the first quadrant)

Question 10.
Find the area bounded by the curve x² = 4y and the line x = 4y – 2
Solution:
Given curve is x² = 4y …(i)
which is an upward parabola with vertex at (0,0) and it is symmetrical about y-axis
Equation of the line is x = 4y – 2 …(ii)
Solving (i) and (ii) simultaneously, we get; (4y – 2)² = 4y
⇒ 16y² – 16y + 4 = 4y
⇒ 4y² – 5y + 1 = 0

Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Solution:
The curve y² = 4x is a parabola as shown in the figure. Axis of the parabola is x-axis. The area of the region bounded by the curve y² = 4x and the line x = 3 is
A = Area of region OPQ = 2 (Area of the region OLQ)

Choose the correct answer in the following Exercises 12 and 13:

Question 12.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
(a) π
(b)
(c)
(d)
Solution:
(a) x² + y² = 4. It is a circle at the centre (0,0) and r=2

Question 13.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
(a) 2
(b)
(c)
(d)
Solution:
(b) y² = 4x is a parabola

Chapter 8 Application of Integrals Exercise 8.2

Question 1.
Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.
Solution:
Area is bounded by the circle 4x² + 4y² = 9 and interior of the parabola x² = 4y.
Putting x² = 4y in x² + y² =
We get 4y + y² =

Question 2.
Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.
Solution:
Given circles are x² + y² = 1 …(i)
and (x – 1)² + y² = 1 …(ii)
Centre of (i) is O (0,0) and radius = 1

Question 3.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.
Solution:
Equation of the parabola is y = x² + 2 or x² = (y – 2)
Its vertex is (0,2) axis is y-axis.
Boundary lines are y = x, x = 0, x = 3.
Graphs of the curve and lines have been shown in the figure.
Area of the region PQRO = Area of the region OAQR-Area of region OAP

Question 4.
Using integration find the area of region bounded by the triangle whose vertices are (-1,0), (1,3) and (3,2).
Solution:
The points A (-1,0), B( 1,3) and C (3,2) are plotted and joined.
Area of ∆ABC = Area of ∆ ABL + Area of trap. BLMC – Area of ∆ACM …(i)
The equation of the line joining the points

Question 5.
Using integration find the area of the triangular region whose sides have the equations y = 2x + 1,y = 3x + 1 and x = 4.
Solution:
The given lines are y = 2x + 1 …(i)
y = 3x + 1 …(ii)
x = 4 …(iii)
Subtract (i) from eq (ii) we get x = 0, Putting x = 0 in eq(i) y = 1
∴ Lines (ii) and (i) intersect at A (0,1) putting x = 4 in eq. (2) =>y = 12 + 1 = 13
The lines (ii) and (iii) intersect at B (4,13) putting x=4ineq. (i):y = 8 + 1 = 9
∴ Lines (i) and (ii); Intersect at C (4,9),

Question 6.
Smaller area bounded by the circle x² + y² = 4 and the line x + y = 2
(a) 2 (π – 2)
(b) π – 2
(c) 2π – 1
(d) 2(π + 2)
Solution:
(b) A circle of radius 2 and centre at O is drawn.The line AB: x + y = 2 is passed through (2,0) and (0,2). Area of the region ACB
= Area of quadrant OAB – Area of ∆OAB …(i)

Question 7.
Area lying between the curves y² = 4x and y = 2x.
(a)
(b)
(c)
(d)
Solution:
(b) The curve is y² = 4x …(1)
and the line is y = 2x …(2)

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